Answer:
Explanation:
Given that,
The radius of loop r = 7cm = 0.07m
Mass of hoop M = 2kg
The hoop is released from rest, the initial velocity is 0m/s
A. Angular speed of the hoop after a descended of h= 80cm = 0.8m
Applying the conservation of energy
Ei + W = Ef
Where,
Ei = initial energy of the system, I.e the initial potential and kinetic energy
Ef = final energy of the system I.e the final potential and kinetic energy
W = 0, since no external force is acting on the systems
Ui + K.Ei + 0 = Uf + K.Ef
System was initially at rest, K.Ei = 0
Uf = 0 at the zero level
Then,
Mgh = ½ •Icm•w² + ½M•Vcm²
Icm = Mr², since it is circular loop
Then,
Mgh = ½ •Mr²•w² + ½M•Vcm²
M cancels out
gh = ½ •r²•w² + ½ Vcm²
Since Vcm = wr
Then, gh = ½ •r²•w² + ½ w²r²
gh = r²w²
w = √(gh/r²)
w = √(9.81 × 0.8/0.07²)
w = 40 rad/s
b. Speed at the center
Since Vcm = wr
Vcm = 40×0.07
Vcm = 2.8 m/s
